Question

Решите неравенство –3x2 – 2x + 5 < 0;

Answer 1

$-3x^2 - 2x + 5 \ \textless \ 0 \\ -3x^2-2x+5 = 0 \\ D=4+60=64 \\ x_{1,2} = \frac{2 \ \pm \ 8}{-6} = \left \{ {{x_1=1} \atop {x_2= -1\frac{2}{3} }} \right. \\ x\in (-\infty;-1\frac{2}{3}) \cup (1;\infty)$

Answer 2

–3x2 – 2x + 5 < 0
3x2 + 2x - 5 > 0
D=4+60=64
x1=(-2-8)/6=-5/3
x2=(-2+8)/6=1
    +                      _                        +
(-5/3)(1)
x∈(-∞;-5/3) U (1;∞)